Dr. Sangeeta Khanna Ph.D 1 · 7. Assertion: Fumaric acid and maleic acid both have two COOH groups,...

$: Dr. Sangeeta Khanna Ph.D 1 · 7. Assertion: Fumaric acid and maleic acid both have two COOH groups, one / \ \ / C C yet their acidic strength is different. Reason: Maleic acid after$
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+2\Organic\Grand Test\+2 Grand Test - VII\+2 Grand Test-VII L-2.docx

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph 2 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+2\Organic\Grand Test\+2 Grand Test - VII\+2 Grand Test-VII L-2.docx

Test Date: 19.11.2017 (Level – 2) Topic: Oxygen Containing Organic Compound

READ INSTRUCTIONS CAREFULLY 1. The test is of 1 hour duration. 2. The maximum marks are 246. 3. This test consists of 48 questions. 4. Keep your mobiles switched off during Test in the Halls.

Section – A (Single Correct Choice Type) Negative Marking [-1]

This Section contains 14 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. 14 × 5 = 70 Marks

1. Reaction of ‘A’ with H2SO4 dil./HgSO4 Gives a compound (B), which can also be obtained from a

reaction of Benzene with acid chloride in the presence of anhydrous AlCl3. The compound (B) when treated with iodine in aq. KOH, yields C and a yellow compound (D). A, B, D are respectively

a. C6H5 – C CH, C6H5 – COOH, C6H6 b. 3

| |

56 CH

O

CHC , C6H5–CCH, C6H5 – COOK

c. C6H5C CH, . 3

| |

56 CH

O

CHC , CHI3 d. C6H6, 3

| |

56 CH

O

CHC , CHI3

C Sol. Formation of (B) from benzene and acid chloride in the presence of anhydrous AlCl3 indicates that it is

a ketone. Formation of yellow compound (D) by reaction of ‘B’ with I2/KOH (Iodoform test) indicates

that ‘B’ is a methyl ketone. Hence ‘B’ is 3

| |

56 CH

O

CHC . Reaction of ‘A’ with H2SO4 dil./HgSO4 to give

ketone indicates that ‘A’ must be C6H5 – C CH.

2. Observe the following reactions

I) II)

(C)

(D)

C6H5 – C CH

(A)

HgSO4

dil. H2SO4

C6H5 – C – CH3 (B)

O

I2/KOH C6H5COOK + CHI3

C6H6 + CH3COCl

Anhyd. AlCl3

CH3

CH3 H3C O HI X(Alcohol) + Y (Alkyl halide)

CH3

CH3

H3C

O HI

W + S

CH3



III) X )1R(

(S)

IV) Y )2R(

(W)

The reagents R1 and R2 can be respectively. a. NaI/Acetone, aq. AgNO3 b. aqueous KOH, HI

c. aq. AgNO3, NaI/Acetone d. HI, aqueous KOH D Sol.

3. What is the product of the following reaction?

)excess(HBr

A B ,O3H

KCN

a. b. c. d. D Sol.

4. An optically active alcohol (P) C6H10O upon catalytic hydrogenation absorbs two moles of hydrogen per mole of (P) and gives a product (Q). The compound (Q) is optically inactive and resistance to oxidation by CrO3. Structures of (P) and (Q) respectively are

a. OHH

CHCH

CCHCHCH,OHH

CH CH

CCHH

CH

C

322

3

|

223

|

|

b. OH

HC

CH

CHC ,OH

HC

CH

CCHC

52

3

52

3

|

|52

|

|

c. CHC

HC

OH

CCH ,OH

HC

HC

CCH

5252

52

|

|3

|

|3

d. 32

|

23

OH

H

|

|52 CHCHH

OHCH

CCHCH ,CHC

HC

CHC

22

B

5.

X = OH Y = CH3CH2 – I S =

I

W = CH3CH2OH

O

Br

COOH

O

O

OH

COOH

O O

OH Br

(B)

CN

OH O O (A)

OH

COOH

OH(3)

HO (2)

HC C (1)

CH2 – NH2(4)



Arrange the hydrogens in the decreasing order of acidity.

a. 1 > 2 > 3 > 4 b. 4 > 3 > 2 > 1 c. 2 > 3 > 1 > 4 d. 2 > 3 > 4 > 1 C 6. Which is the major product of the following reaction?

a. b.

c. 3

||

22

||

CHNH

O

CCHCH

O

CHO d. 3

||

22

||

3 CHNH

O

CCHCH

O

CNHCH

A

7. In the given reaction: ]X[COOHCHO3H )ii(

)excess( Li3CH )i(3

[X] is:

a. CH3COOLi b. CH3COOCH3 c. 3

3

|

|3 CH

OH

CH

CCH d. 3

||

3 CH

O

CCH

D Sol. Acid reacts with excess of alkyl lithium to give ketone

O

CH

CCHLiOH2OLi

LiO

CH

CCHLiO

O

CCHCHOH

O

CCH

33

33

|3

HOH|

|3

LiCH||

34LiCH

||

3

8. In the given reaction, the product is:

a. b.

c. d. C

O

O

O + CH3 – NH2

product

O

O

N – CH3 O

O

N – CH3

O

CH2 – COOH

COOH

HOOC

O

CH2 – COOH

HOOC O

HOOC

CH3 COOH

O

O = C

CH3

O



Sol. 9. Which pair of reactants compounds may be used to make given acetal?

a. b.

c. d.

D

Sol.

10. An organic compound of molecular formula C5H10O2 gives the following properties.

(i) It evolves effervescence with NaHCO3 solution (ii) Its sodium salt when fused with soda lime gives neobutane. (iii) Its calcium salt when distilled gives di-ter-butyl ketone. The structure of the original compound is a. (CH3)3C – COOH b. CH3 – CH2 – CH2 – CH2 – COOH

c. COOHCHH

CH

CCH 2

3

|3 d. COOHH

CH

CH

CH

CCH

3

|

3

|3

A

Sol. etanneobu

33lime Soda

NaOH

acid oictanbuter33 CHCHCOOHCCH

(i) Ca(OH)2 (ii) distil

ketone butyltertdi

3333 CHCCOCCH

Since the compound liberates CO2 with NaHCO3 solution, it is an acid. It forms only one monochloro product, shows that it contains only one type of alkyl group and it may be (CH3)3C – COOH. This is proved by its conversion to neobutane and di-ter-butyl ketone. In each of the following questions three statements are given. Mark the correct answer as per following instructions

(a) All are correct (b) If all are wrong (c) If II and III is correct (d) If I and II are correct

O

HOOC

Behaves as 1, 7-dicarboxylic acid COOH

CH2 – COOH

O

CH2 – COOH

HOOC O

O = C

+ H2O + CO2

O O

O

+

CH2 –OH

CH2 –OH

O

+ OH

OH

O

+

OH

OH

O

+ OH

OH

O

+

OH OH

H+

O O



11. I. Hydrolysis of methylisocyanide produces formic acid. II. Oxidation of benzyl chloride with Pb(NO3)2 produces benzaldehyde III. Hydrogenation of Benzoyl chloride in the presence of Pd-BaSO4 produces benzaldehyde.

a. (a) b. (b) c. (c) d. (d) A 12. I. Acetone gives tertiary butyl alcohol on reduction. II. Acetone gives acetic acid on drastic oxidation. III. Acetone undergoes bimolecular reduction with Mg –Hg/H2O.

a. (a) b. (b) c. (c) d. (d) C NH 13. I. HO – C is imide form of urea. NH2 II. Conc. H2SO4 and formic acid react to form carbon monoxides gas III. pKa of chloroacetic acid is higher than that of benzoic acid.

a. (a) b. (b) c. (c) d. (d) D 14. I. Tartaric acid does not contain – COOH group at all

II. Crotonic acid does not decolourise bromine dissolved in CS2. III. Alkane with same number of C atoms cannot be produced from carboxylic acid.

a. (a) b. (b) c. (c) d. (d) B

SECTION – B (ASSERTION & REASON)

This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and

D) out of which ONLY ONE is correct. (10 × 5 = 50 Marks)

(a) Mark A if both A and R are correct and R is the correct reason of A. (b) Mark B if both A and R are correct and R is not the correct reason of A. (c) Mark C if A is correct and R is wrong.

(d) Mark D if A is wrong and R is correct. 1. Assertion : Boiling points of esters are higher than corresponding isomeric carboxylic acid . Reason : Acid molecule dimerise in non–aqueous solutions. a. (A) b. (B) c. (C) d. (D) D 2. Assertion: Relative reactivity of acid derivatives to nucleophilic acyl substitution is :

acid chloride > anhydride > amide > ester Reason: Weaker the conjugate base, better the leaving group, more the reactive it is.

a. (A) b. (B) c. (C) d. (D) D 3. Assertion: Iodoform is obtained by the reaction of acetone with hypoiodite and not with iodide.

Reason: OI– is an oxidizing agent as well as an iodinating agent.

a. (A) b. (B) c. (C) d. (D) B

4. Statement – 1: Cyclopentane-1,2-dione has more stable keto form than butane-2,3-dione. Statement – 2: In butane-2,3-dione keto form have two keto groups at anti position a. (A) b. (B) c. (C) d. (D) D



5. Assertion: Diisopropyl ketone is more reactive than acetone. Reason: Acidic medium makes the carbonyl group more susceptible to nucleophilic attack. a. (A) b. (B) c. (C) d. (D) D

Sol. OH

Nu|C

\

/

R

ROHC

\

/

R

ROC

\/

R

R

ΘNuH

6. Assertion: Hydroxy ketones are not directly used in Grignard reaction.] Reason: Grignard reagents react with hydroxyl group.

a. (A) b. (B) c. (C) d. (D) A

7. Assertion: Fumaric acid and maleic acid both have two COOH groups, one /\

\/

CC yet their acidic

strength is different. Reason: Maleic acid after the loss of H+ is stabilized due to intramolecular H-bonding, so ease of release of H+ is easier in maleic acid as compare to fumaric acid.

a. (A) b. (B) c. (C) d. (D) A

8. Assertion: -keto butyric acid is esterified faster than butyric acid. Reason: Groups increasing the intensity of +ve charge on acidic C-atom to increase the reactivity

towards esterification OC\/

group being e– withdrawing increases +ve charge so esterified faster

than butyric acid. a. (A) b. (B) c. (C) d. (D)

A

9. Assertion: Group like – NO2, – CN,

O

C||

attached at - C–atom to – COOH group makes decarboxylation faster.

Reason: Strong e– withdrawing group at - C-atom facilitates the decarboxylation due to stable carbocation.

a. (A) b. (B) c. (C) d. (D) C

10. Assertion : m–Nitrobenzoic acid is less stronger acid as compared to p–nitrobenzoic acid. Reason : Acetic acid is stronger acid than benzoic acid but formic acid is still stronger than both.

a. (A) b. (B) c. (C) d. (D) C

SECTION – C (Paragraph Type)

This Section contains 2 paragraph. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 6 × 5 = 30 Marks

Passage – 1

Aldehydes and ketones have sp2 hybridised carbon and they undergo nucleophilic addition reactions.

Aldehydes are more reactive than ketones. Aldehydes and ketones having at least one -hydrogen

undergo aldol condensation, while aldehydes with no -hydrogen undergo Cannizzaro reaction. Ketones do not undergo Cannizzaro reaction



1. The products of the action of concentrated alkali on a mixture of formaldehyde and benzaldehyde are

a. sodium benzoate, methanol b. benzylalcohol, sodium fomate c. sodium benzoate, sodium formate d. benzylalcohol, methanol B

2. The product of reaction between benzaldehyde and malonic ester

52

522

HCOOC

HCOOCCH

/\

in presence of

pyridine followed by acid hydrolysis and heating is

a. C6H5 – CH(OH) – CH2 – COOH b. C6H5 – CH = CH – COOH c. C6H5 – CH2 – CH2 – COOH d. C6H5 – CH = CH – C6H5 B

3. The product/s of reaction in alkaline medium between acetophenone and benzaldehyde is/are

a. C6H5COOH and C6H5CHOCH3 b. C6H5 – CH = CH – CO – C6H5 c. C6H5 – CO – CH2 – CH2 – C6H5 d. C6H5CH2OH and C6H5 – CHOH – CH3 B

Passage -2 Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compounds S. On acidification and heating, S gives the product shown below:

4. The compounds P and Q respectively are

a. b. c. d.

B 5. The compound R is

a. b.

OH

O

O

H3C

H3C

CH

CH3

C

H

H3C and C

H

H3C

O

O

H3C

CH

CH3

CH2

C

H

O

and C

H

H3C

O

H3C

C H

C

O

H3C

CH2

OH

H3C

C H

C

O

H3C

CH

OH

H3C

CH

CH3

C

H

H3C and C

H

O

O

H

H3C

CH

CH3

CH2

C

H

O

and C

H

O

H



c. d. A 6. The compound S is

a. b.

c. d.

D Sol.

CH C

H3C

CH3

CH

O

H

CH2

OH

CH C

H3C

CH3

CH

O

H

CH

OH H3C

CH C

H3C

CH3

CH

O

H

CH2

CN

C

C

H3C

O

H

CH2

CN

H3C

CH

H3C

CH3

CH

CH2

OH

CH

OH

CN

C

H3C

CH2

H3C

CN

OH

OH

CH

H3C C CH

CH3 OH

O O

H3C C CH

CH3 OH

O OH

HO

H+

H3C C CH

CH3 OH

OH

CN

(S)

HCN

H3C C CHO

CH3

OH

(R)

aq.K2CO3 H3C CH CHO + H C H

O

CH3

(Q)

(P)



SECTION – D (More than One Answer Type) No Negative Marking

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D) out of which One or More than one answer may be correct. 8 × 5 = 40 Marks

1. The ether , when treated with HI produces

a. b. c. d. A, D 2. Which of the following undergo decarboxylation upon heating?

(A) (B) (C) (D) a. A b. B c. C d. D A, D 3. Which of the following process is involved in Reimer Tiemann’s reaction

a. Nucleophilic substitution b. Carbene intermediate c. Rearrangement d. Electrophilic substitution A, B, C, D

4. Formic acid and acetic acid can be distinguished by the action of

a. conc. H2SO4 b. Tollen’s reagent c. Fehling’s solution d. NaHCO3

A,B,C Sol. Both give CO2 with NaHCO3

Formic acid Acetic acid

Conc. H2SO4 HCOOH42SOH

.conc CO + H2O No effect

Tollen’s Reagent HCOOH + Ag2O

ppt. Black

Ag2 + CO2 + H2O No effect

Fehling’s solution HCOOH + 2CuO

ppt. dRe2OCu + CO2 + H2O

No effect

5. Which of the following does form a stable hydrate by the addition of H2O?

a. b. c. d.

A,B,C

O

CH2I CH2OH I OH

O

CO2H CO2H

CO2H COOH

COOH

OH

COOH

Ph – C – C – C – Ph

O O O

O

O

O

O

O



6. Which of the following can be the product of following reaction

a. b. c. d. A,C,D Sol.

7.

a. (X) is b. (Y) is c.(X) is d. (Y) is A, D

OH

N

HO

OH

N N

N

N N O

O

O

N

N N O

O

O

OH

HN

HO

OH

NH NH

CH2OH

OH

CH2OH

O

C – OCH3

OH

O

C – OCH3

OH

O

C – OCH3

O

O

LiAlH4 (X)

NaBH4

(Y)

NH2OH

OH

OH

HO

NH2OH

O

Ketoform

NHOH

OH HO

HONH

NHOH HO

NHOH

NHOH

HONH

NOH

NOH

HON

enolic

form

NO

NO

ON

Ketoform

O

O

OH

OH HO



8. Which of the following reagent(s) can be used to convert amide into a primary amine with one carbon atom less than amide?

a. Br2 + NaOH b. NaOBr c. P2O5 d. Br2 + Na2CO3 A, B, D

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 question. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 2 = 16 Marks

1. Match each of the compounds in Column I with its characteristic reaction(s) in Column II. (One or

More than One Match) Column I Column II

(a) CH3CH2CH2CN (p) Reduction with Pd-C/H2 (b) CH3CH2OCOCH3 (q) Reduction with SnCl2/HCl (c) CH3 – CH = CH – CH2OH (r) Development of foul smell on treatment with chloroform

and alcoholic KOH (d) CH3CH2CH2CH2NH2 (s) Reduction with diisobutyl aluminium hydride (DIBAL-H) (t) Alkaline hydrolysis

Sol. A p, q, s, t; B s, t ; C p; D r 2. Match compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. (One or

More than One Match). Column – I Column– II (A)

Θ

32 ClHNNH

(p) Compound give effervescence with NaHCO3

(B)

(q)

gives positive FeCl3 test

(C)

(r)

gives white precipitate with AgNO3

(D)

(s)

reacts with aldehydes to form the corresponding hydrazone derivative

Sol.A – r, s ; B – p, q ; C – q, r ; D – s

SECTION – F (Integer Type) No Negative Marking

This Section contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9. (8 × 5 = 40 Marks)

1. COOHHCMgBrHC 1n2nO2H)ii(

2CO)i(52 . The value of n is

Sol. 2 CH3COCl + CH3COONa CH3 – CO – O – COCH3 2. The number of moles of ethanoyl chloride which produces 10.2 g of ethanoic anhydrie on reaction with

sodium acetate is 10-x, X is: Sol. 1 3. How many different Grignard reagents having single Bromine can give n-butane as product (excluding

stereisomer), when react with C2H5OH Sol. 2

CH3 – CH2 – CH2 – CH2 – MgBr, 3|

23 CHH

MgBr

CCHCH

Above Grignard reagent when reacts with ethanol, normal butane as a product. 4. How many isomeric ketones can be formulated by C6H12O?

HO NH3 I

COOH

HO NH3 Cl

O2N NH – NH3Br

NO2



Sol. 7

;C

O

CCCCC||

CCC

O

CCC ||

;

/d

|||

C

O

C

C

CCC

; CC

C

C

CC

O|||

| ; CC

O

CC\/

C

C

||

5.

Number of moles (x) of Grignard reagent consumed in the above reaction is: Sol. 4 6. An organic compound ‘A’ having molecular formula C5H10O6 on acetylation forms a compound having

molecular weight 334. Number of hydroxy groups (-OH) in compound ‘A’ is: Sol. 4

On acetylation, ‘OH’ group get changed into – OCOCH3 means per ‘OH’ group molecular weight increases by ‘42’.

Molecular wt. of ‘A’ C5H10O6 166. Molecular wt. after acetylation = 334 increases in molecular wt. = 334 – 166 = 168.

Number of ‘OH’ groups = 442

168

7. Complete the following reactions and find out number of oxygen atom present in compound (D)D i)

Sol. 3 8. An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and compound ‘C’.

Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMnO4 also gives ‘B’. ‘B’ on heating with Ca(OH)2 gives ‘E’. E does not give Tollen’s test and does not reduce Fehling’s solution but form a 2,4-dinitrophenyl hydrazone. How many carbon are present in product (E).

Sol. 3

CO2CH3

CO2CH3

x CH3MgI

H+

C(CH3)2

C(CH3)2

OH

; Dimethyl phthalate

OH

O

+ HC CNa (A) HgSO4

H2SO4 (B)

(C)

O3 (D)

A =

OH

CH B =

OH CH3

O

C =

O

CH3

D = CHO

O

O

CH3

HOH/Zn

D = H5C2 – OH E = H3C

O

CH3

B = H3C

OH

O

Ca(OH)2

(CH3COO)2Ca

Ester

+ CaCO3

A =

O

H3C

O

H3C

O

C =

O

H3C

O

H5C2

Dr. Sangeeta Khanna Ph.D 1 · 7. Assertion: Fumaric acid and maleic acid both have two COOH groups,...

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Transcript of Dr. Sangeeta Khanna Ph.D 1 · 7. Assertion: Fumaric acid and maleic acid both have two COOH groups,...